[EDIT]
0) In the EPR paradox (in fact the CHSH version), based on the hypothesis of local realism , the apparent impossibility of measuring $x$,$z$ polarizations for an entangled state, means that the choice would be between :
a)some interaction exists between the particles, even though they were separated
b) (realism) the information about the outcome of all possible measurements was already present in both particles (hidden parameters).
The first possibility seems incompatible with locality, so Einstein chooses the second possibility (realism), and says that quantum mechanic was incomplete (because we have to add these hidden parameters to the description of each particle).
But experience has showed that Quantum Mechanics violate Bell's inequalities (local realism). So we have to choose between get rid of locality, or get rid of realism. The correct choice is get rid of realism, that is, you cannot consider the 2 particles individually, you have to consider the entangled particles as a whole, you cannot consider the 2 particles separately . This does not mean that Quantum Mechanics violates locality, Quantum Mechanics respects locality. For instance, with entangled particles, it is not possible to send information instantly. Simply, the quantum correlations are stronger than classical correlations.
1) So one first fundamental idea is that the entangled 2-quit state is a whole, and cannot be divided into more little units.
You cannot separate the 2 qbits and the operators acting on them.
2) Alice and Bob can freely choose the orientation of their measurement apparatus, and they always obtain an outcome (your video is wrong about this) which maybe 0 or 1.
3) The fact that quantum mechanics does not respect realism can be seen in the mathematical formalism. For instance, measurements are operators like 2*2 Pauli Matrices. A state is a 2-dimensional complex vector. So applying a measurement to a state, is the same thing that looking at the result of a matrix applying to a vector, for instance :
$$\sigma_x |0\rangle = |1\rangle$$
You see that the measurement change the state, the state after the measurement is not the same that the state before the measurement, so there is no more realism
4) In the case of 2 entangled qbits quantum mechanics, you have to use the formalism of tensorial products. If Alice choose a z-axis measurement, and Bob a x-axis measurement, that means that the measurement operator is :
$$\sigma_z \otimes \sigma_x$$
where $\sigma_z$, $\sigma_x$, are
The above expression is a tensorial product of operators, here it is the tensorial product of the matrices $\sigma_z$ and $\sigma_x$.
It works like this, suppose a separable state $|a\rangle \otimes ~|b\rangle$, then :
$$(\sigma_z \otimes \sigma_x) (|a\rangle \otimes ~|b\rangle) = (\sigma_z|a\rangle) \otimes ~ (\sigma_x|b\rangle)$$
For instance, with your entangled state $|\Psi\rangle= \frac{1}{\sqrt{2}} (|0\rangle \otimes |0\rangle + |1\rangle \otimes |1\rangle )$, it gives :
$$(\sigma_z \otimes \sigma_x)|\Psi\rangle = \frac{1}{\sqrt{2}}(\sigma_z \otimes \sigma_x) (|0\rangle \otimes |0\rangle + |1\rangle \otimes |1\rangle ))$$
$$(\sigma_z \otimes \sigma_x)|\Psi\rangle = \frac{1}{\sqrt{2}}((\sigma_z|0\rangle \otimes ~ (\sigma_x|0\rangle\rangle) + (\sigma_z|1\rangle) \otimes ~ (\sigma_x|1\rangle))$$
$$(\sigma_z \otimes \sigma_x)|\Psi\rangle = \frac{1}{\sqrt{2}}((|0\rangle \otimes ~ (|1\rangle) - (|1\rangle) \otimes ~ (|0\rangle))$$
You see again that the measurement changes the state, so there is no more realism here too.
Best Answer
The problem with this sort of scheme is that Alice has no control over the results of her measurements, since those are random. This means that she can control which basis Bob's spin is projected on, but she cannot control which of the basis states gets chosen. Bob will then see a random mix of results which turns out to contain no trace of what Alice was trying to communicate.
To make this more precise, consider the standard case where they share a Bell triplet state $$ \newcommand{\up}{|\!\uparrow\rangle}\newcommand{\down}{|\!\downarrow\rangle} \newcommand{\plus}{|+\rangle}\newcommand{\minus}{|-\rangle} |\Phi\rangle=\up\up+\down\down $$ (ignoring normalization) at the start of the protocol, which they use as a resource state. Alice can choose to measure along the $z$ direction, in the basis $\{\up,\down\}$, or along the $x$ direction, in the basis $\{\plus=\tfrac1{\sqrt{2}}(\up+\down),\minus=\tfrac1{\sqrt{2}}(\up-\down)\}$. Because of the nice properties of the triplet state, whatever state Alice's qubit is projected on (in these two bases) will be identically replicated in Bob's qubit. Both states of either basis come up with equal probabilities.
Alice's only choice in this scheme is what basis she measures in, and she can transmit one bit of information if she can engineer a situation where Bob can determine that basis. Assume, if you want to, that she can repeat this protocol $n$ times, with $n$ possibly greater than one, to help ensure the information gets there.
Suppose, then, that Alice chose to measure in the $z$ direction. How can Bob determine this fact? To put this more explicitly, how can he determine that Alice didn't measure in the $x$ direction? His problem, then, is to determine whether his ensemble of $n$ qubits is in a random mix of $\up$s and $\down$s, or in a random mix of $\plus$s and $\minus$s.
Unfortunately, this is impossible to do. If he measures in the $z$ direction, he will get fifty/fifty results if Alice is sending $\up$s and $\down$s, but he would also get fifty/fifty chances from each $\plus$ or $\minus$, and therefore from the whole set, if Alice had measured in the $x$ direction. Regardless of what basis Alice chose or what basis he himself measures in, both situations look exactly the same to Bob.